Let $A$ be an $n \times n $ matrix.
Is my proof of the above correct?
Proof
Let the columns of $A$ be denoted by $a_1, a_2, …, a_n \implies A = [a_1, a_2, …, a_n]$.
$A^2 = 0 \iff A[a_1, a_2, …, a_n] = 0 \iff [Aa_1, Aa_2, …, Aa_n] = 0$
$\iff Aa_1 = Aa_2 =\ldots= Aa_n = 0$
$\iff C(A) \subseteq N(A)$
Also, I have proved up to this point that the column space of A is a subset of its null space. How can I prove that it is furthermore a proper subset?
Best Answer
My previous answer was completely wrong actually, and I'm sorry about that. In fact there is no reason why $C(A)=N(A)$ cannot happen. Here is an example: $$ A = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} $$ Clearly $A^2=0$. Now $C(A)=\mathrm{span}(1,0)=N(A)$.
The reason why my previous answer fails is that it is true that $A$ sends $\mathrm{span}(a_1, \cdots, a_n)$ to zero, but this doesn't mean $A=0$ identically (as I falsely argued before). This is because not all the vectors in the $n$-dimensional vector space are a linear combination of $a_1, \cdots, a_n$. This is only true if $a_1, \cdots, a_n$ are linearly independent, in which case $A$ is invertible and $A^2=0$ cannot happen.
Edit: One thing can be said though, since $C(A)\subseteq N(A)$, and $\dim C(A)=\mathrm{rank} A$, then $\mathrm{rank}A\leq \mathrm{null}A$. Therefore $2\mathrm{rank}A\leq n$. If $n$ is odd, then we cannot have $C(A)=N(A)$, since then $2\mathrm{rank}A = n$, a contradiction. But for even $n$, the equality can happen.